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(B) The root mean square (r.m.s.) velocity of a gas molecule is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$.Here, $R$ is the universal gas

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$320$

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(B) The root mean square (r.m.s.) velocity of a gas molecule is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$.Here, $R$ is the universal gas constant, $T$ is the absolute temperature, and $M$ is the molar mass of the gas.For hydrogen $(H_2)$, the molar mass $M_{H_2} = 2 	imes 10^{-3} \ kg/mol$ and temperature $T_{H_2} = 20 \ K$.For oxygen $(O_2)$, the molar mass $M_{O_2} = 32 	imes 10^{-3} \ kg/mol$.We are given that the r.m.s. velocities are equal: $v_{rms(H_2)} = v_{rms(O_2)}$.Therefore, $\sqrt{\frac{3RT_{H_2}}{M_{H_2}}} = \sqrt{\frac{3RT_{O_2}}{M_{O_2}}}$.Squaring both sides and simplifying: $\frac{T_{H_2}}{M_{H_2}} = \frac{T_{O_2}}{M_{O_2}}$.Substituting the values: $\frac{20}{2} = \frac{T_{O_2}}{32}$.$10 = \frac{T_{O_2}}{32} \implies T_{O_2} = 320 \ K$.

At what temperature does an oxygen molecule have the same r.m.s. velocity as a hydrogen molecule at $20 \ K$ (in $K$)?

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